Editing Thales's theorem (section)

===First proof===
The following facts are used: the [[sum of angles of a triangle|sum of the angles]] in a [[triangle]] is equal to 180° and the base angles of an [[isosceles triangle]] are equal.
{{Gallery
 |File:Animated illustration of thales theorem.gif|Provided {{mvar|{{overline|AC}}}} is a [[diameter]], angle at {{mvar|B}} is constant [[right angle|right]] (90°).
 |File:Thales' Theorem.svg|Figure for the proof.
}}

Since {{math|1=''{{overline|OA}}'' = ''{{overline|OB}}'' = ''{{overline|OC}}''}}, {{math|△''OBA''}} and {{math|△''OBC''}} are isosceles triangles, and by the equality of the base angles of an isosceles triangle, {{math|1=∠ ''OBC'' = ∠ ''OCB''}} and {{math|1=∠ ''OBA'' = ∠ ''OAB''}}.

Let {{math|1=''α'' = ∠ ''BAO''}} and {{math|1=''β'' = ∠ ''OBC''}}. The three internal angles of the {{math|∆''ABC''}} triangle are {{mvar|α}}, {{math|(''α'' + ''β'')}}, and {{mvar|β}}. Since the sum of the angles of a triangle is equal to 180°, we have

:<math>\begin{align}
\alpha+ (\alpha + \beta) + \beta &= 180^\circ \\
2\alpha + 2\beta &= 180^\circ \\
2( \alpha + \beta ) &= 180^\circ \\
\therefore \alpha + \beta &= 90^\circ.
\end{align}</math>

[[Q.E.D.]]